CBSE Class 11 Mathematics Straight Lines Competitive Exam MCQs Set 03

Answer: (d) 4

Explanation:
1. Slope of the line through (4, 3) and (2, k) = \( \frac{k - 3}{2 - 4} = \frac{k - 3}{-2} \).
2. For perpendicularity with slope 2: \( \frac{k - 3}{-2} = -\frac{1}{2} \Rightarrow k - 3 = 1 \Rightarrow k = 4 \).

Teacher's Note:
1. Use m₁m₂ = −1 to get the required slope first.
2. Then equate it to the two-point slope and solve for the unknown.

Answer: (c) 150°

Explanation:
1. Slope = \( \frac{\sqrt{3} - 0}{-2 - 1} = -\frac{1}{\sqrt{3}} \).
2. tan θ = \( -\frac{1}{\sqrt{3}} \) with 0° ≤ θ < 180° gives θ = 150°.

Teacher's Note:
1. A negative slope means the inclination is an obtuse angle.
2. tan 150° = −tan 30° = −1/√3.

Answer: (b) 3x − 2y = 1

Explanation:
1. Multiply the equations: 8x − 6y = 2 and 15x − 6y = 9; subtracting, 7x = 7 ⇒ x = 1, y = 1.
2. Slope of 2y − 3x + 2 = 0 is \( \frac{3}{2} \).
3. Line: \( y - 1 = \frac{3}{2}(x - 1) \Rightarrow 2y - 2 = 3x - 3 \Rightarrow 3x - 2y = 1 \).

Teacher's Note:
1. Make the coefficients of one variable equal before subtracting.
2. A line parallel to 2y − 3x + 2 = 0 can also be written as 3x − 2y = λ.

Answer: (a) x − y − 10 = 0

Explanation:
1. Slope = tan 45° = 1.
2. Equation: y + 6 = 1(x − 4) ⇒ x − y − 10 = 0.

Teacher's Note:
1. The slope of a line is the tangent of its inclination.
2. Check: 4 − (−6) − 10 = 0.

Answer: (b) 5x + 4y = 0

Explanation:
1. From x + 2y = 10, x = 10 − 2y. Then 2(10 − 2y) + y = −5 ⇒ y = \( \frac{25}{3} \), x = \( -\frac{20}{3} \).
2. Line through the origin and \( \left(-\frac{20}{3}, \frac{25}{3}\right) \): slope = \( -\frac{25}{20} = -\frac{5}{4} \).
3. So 4y = −5x, i.e., 5x + 4y = 0.

Teacher's Note:
1. A line through the origin is y = mx, so only the ratio y/x of the second point is needed.
2. Check: 5(−20/3) + 4(25/3) = 0.

Answer: (a) x cos θ − y sin θ = a cos 2θ

Explanation:
1. Slope of x sec θ + y cosec θ = a is \( -\frac{\sec\theta}{\csc\theta} = -\tan\theta \), so the perpendicular slope is cot θ.
2. Line: \( y - a\sin^3\theta = \frac{\cos\theta}{\sin\theta}(x - a\cos^3\theta) \).
3. Multiplying by sin θ: y sin θ − a sin⁴ θ = x cos θ − a cos⁴ θ.
4. So x cos θ − y sin θ = a(cos⁴ θ − sin⁴ θ) = a(cos² θ − sin² θ) = a cos 2θ.

Teacher's Note:
1. cos⁴ θ − sin⁴ θ factorises as (cos² θ − sin² θ)(cos² θ + sin² θ) = cos 2θ.
2. Treat sec θ and cosec θ as ordinary coefficients when finding the slope.

Answer: (c) 4x + 3y = 15

Explanation:
1. Mid-point = (3, 1). Slope of the segment = \( \frac{-2 - 4}{-1 - 7} = \frac{3}{4} \), so the perpendicular slope is \( -\frac{4}{3} \).
2. Equation: \( y - 1 = -\frac{4}{3}(x - 3) \Rightarrow 3y - 3 = -4x + 12 \Rightarrow 4x + 3y = 15 \).

Teacher's Note:
1. The right (perpendicular) bisector passes through the mid-point and is perpendicular to the segment.
2. Every point on it is equidistant from the two end points, which gives a quick check.

Answer: (b) y − 1 = ±1(x − 1)

Explanation:
1. 4x − 3y = 1 and 4x − 10y = −6; subtracting, 7y = 7 ⇒ y = 1, x = 1.
2. Lines equally inclined to the axes have slope ±1.
3. Hence y − 1 = ±1(x − 1).

Teacher's Note:
1. A line equally inclined to both axes makes 45° or 135° with the x-axis, so its slope is 1 or −1.
2. Find the common point first, then apply point-slope form.

Answer: (d) 3y = 4x + 2

Explanation:
1. Perpendicular slope = \( \frac{4}{3} \).
2. \( y - 2 = \frac{4}{3}(x - 1) \Rightarrow 3y - 6 = 4x - 4 \Rightarrow 3y = 4x + 2 \).

Teacher's Note:
1. Slope of 3x + 4y + 5 = 0 is −3/4, whose negative reciprocal is 4/3.
2. Check: 3(2) = 6 = 4(1) + 2.

Answer: (b) 2x − y + 4 = 0

Explanation:
1. Slope = \( \frac{10 + 6}{3 + 5} = 2 \).
2. y − 10 = 2(x − 3) ⇒ y = 2x + 4 ⇒ 2x − y + 4 = 0.

Teacher's Note:
1. Use the two-point form: find the slope, then use either point.
2. Check with the other point: 2(−5) − (−6) + 4 = 0.

Answer: (d) 2x + y − 4 = 0

Explanation:
1. If the line meets the axes at (a, 0) and (0, b), the mid-point is (a/2, b/2) = (1, 2), so a = 2, b = 4.
2. Equation: \( \frac{x}{2} + \frac{y}{4} = 1 \Rightarrow 2x + y - 4 = 0 \).

Teacher's Note:
1. When (h, k) is the mid-point of the intercept, the intercepts are 2h and 2k.
2. This gives the standard result x/h + y/k = 2.

Answer: (a) 5x + 3y + 8 = 0

Explanation:
1. 5x − 6y = 1 and 9x + 6y = −15; adding, 14x = −14 ⇒ x = −1, y = −1.
2. A line perpendicular to 3x − 5y + 11 = 0 is 5x + 3y + λ = 0.
3. Through (−1, −1): −5 − 3 + λ = 0 ⇒ λ = 8, so 5x + 3y + 8 = 0.

Teacher's Note:
1. Perpendicular to ax + by + c = 0 is bx − ay + λ = 0; here that is −5x − 3y + λ = 0, i.e., 5x + 3y + λ′ = 0.
2. Multiply equations so that one variable cancels on adding.

Answer: (a) 3x − y − 1 = 0

Explanation:
1. The diagonal through (1, 2) also passes through the opposite vertex (3, 8).
2. Slope = \( \frac{8 - 2}{3 - 1} = 3 \); y − 2 = 3(x − 1) ⇒ 3x − y − 1 = 0.

Teacher's Note:
1. A diagonal joins opposite vertices, so the two given points lie on the same diagonal.
2. The other diagonal is the perpendicular bisector of this one.

Answer: (a) In A.P.

Explanation:
1. (1, −2) lies on the line ⇒ a − 2b + c = 0.
2. So 2b = a + c, which means a, b, c are in A.P.

Teacher's Note:
1. Three numbers a, b, c are in A.P. exactly when 2b = a + c.
2. Substituting the fixed point into the line gives the required relation immediately.

Answer: (d) 4x + 3y = 0

Explanation:
1. y = x − 7; then x − 7 + 2x − 2 = 0 ⇒ x = 3, y = −4.
2. Line through (0, 0) and (3, −4): \( y = -\frac{4}{3}x \Rightarrow 4x + 3y = 0 \).

Teacher's Note:
1. A line through the origin and (h, k) is kx − hy = 0.
2. Here that gives −4x − 3y = 0, i.e., 4x + 3y = 0.

Answer: (c) x + y + 5 = 0

Explanation:
1. Slope = tan 135° = −1 and the y-intercept is c = −5.
2. y = −x − 5 ⇒ x + y + 5 = 0.

Teacher's Note:
1. tan 135° = −tan 45° = −1.
2. Use the slope-intercept form y = mx + c directly.

Answer: (b) −1

Explanation:
1. x + 2y = 1 and 4x − 2y = −6; adding, 5x = −5 ⇒ x = −1, y = 1.
2. y = mx passes through (−1, 1) ⇒ m = \( \frac{1}{-1} = -1 \).

Teacher's Note:
1. The common point of the two given lines must lie on y = mx.
2. For a line through the origin, m is simply y/x of that point.

Answer: (a) 5x + 3y + 1 = 0

Explanation:
1. A line perpendicular to 3x − 5y + 7 = 0 is 5x + 3y + λ = 0.
2. Through (1, −2): 5 − 6 + λ = 0 ⇒ λ = 1, so 5x + 3y + 1 = 0.

Teacher's Note:
1. Interchanging the coefficients of x and y and changing one sign gives a perpendicular line.
2. Check: slopes 3/5 and −5/3 multiply to −1.

Answer: (a) y = d

Explanation:
1. x = c is a vertical line (parallel to the y-axis).
2. Any line perpendicular to it is horizontal, of the form y = d.

Teacher's Note:
1. Vertical lines have the form x = constant and horizontal lines y = constant.
2. x = 0 is itself vertical, so it is parallel, not perpendicular, to x = c.

Answer: (d) 3π/4

Explanation:
1. Mid-point of (4, −5) and (−2, 9) is (1, 2).
2. Slope of the line through (−3, 6) and (1, 2) = \( \frac{2 - 6}{1 + 3} = -1 \).
3. tan θ = −1 ⇒ θ = \( \frac{3\pi}{4} \).

Teacher's Note:
1. Inclination is measured anticlockwise from the positive x-axis, so it lies in [0, π).
2. A slope of −1 corresponds to 135°, i.e., 3π/4.

Answer: (b) 2x + 5y = 20

Explanation:
1. The intercepts are 2 × 5 = 10 and 2 × 2 = 4.
2. \( \frac{x}{10} + \frac{y}{4} = 1 \Rightarrow 2x + 5y = 20 \).

Teacher's Note:
1. The mid-point of (a, 0) and (0, b) is (a/2, b/2).
2. Equivalently use x/h + y/k = 2 with (h, k) = (5, 2).

Answer: (a) 2x + 3y − 5 = 0

Explanation:
1. Parallel line: 2x + 3y + λ = 0.
2. Through (1, 1): 2 + 3 + λ = 0 ⇒ λ = −5, so 2x + 3y − 5 = 0.

Teacher's Note:
1. Keep the coefficients of x and y unchanged for a parallel line.
2. Find the constant by substituting the given point.

Answer: (a) 5x − y = 0

Explanation:
1. Adding the equations: 3x − 1 = 0 ⇒ x = \( \frac{1}{3} \), y = \( \frac{5}{3} \).
2. Line through the origin: slope = \( \frac{5/3}{1/3} = 5 \), so y = 5x, i.e., 5x − y = 0.

Teacher's Note:
1. Adding the equations eliminates y because its coefficients are +1 and −1.
2. For a line through the origin only the ratio y/x of the other point matters.

Answer: (b) a + b + c = 0

Explanation:
1. Coincident lines need \( \frac{b^3 - c^3}{b - c} = \frac{c^3 - a^3}{c - a} = \frac{a^3 - b^3}{a - b} \).
2. That is b² + bc + c² = c² + ca + a² = a² + ab + b².
3. From the first equality: b² − a² + c(b − a) = 0 ⇒ (b − a)(a + b + c) = 0.
4. Since a ≠ b, we get a + b + c = 0 (the other equality gives the same condition).

Teacher's Note:
1. Two lines are the same when all three ratios of corresponding coefficients are equal.
2. x³ − y³ = (x − y)(x² + xy + y²) lets each ratio simplify.

Answer: (a) x = 2

Explanation:
1. x = 3y − 1; then 2(3y − 1) + 5y − 9 = 0 ⇒ y = 1, x = 2. The lines meet at (2, 1).
2. Infinite slope means the line is vertical: x = 2, which is indeed at a distance 2 from the origin.

Teacher's Note:
1. A vertical line x = k is at a distance |k| from the origin.
2. Only the x-coordinate of the common point is needed for a vertical line.

Answer: (a) y = 3x − 9

Explanation:
1. Using y = m(x − d) with m = 3 and x-intercept d = 3:
2. y = 3(x − 3) = 3x − 9.

Teacher's Note:
1. A line with slope m and x-intercept d is y = m(x − d).
2. Check: y = 0 gives x = 3.

Answer: (c) x − y − 1 = 0, x + y + 1 = 0

Explanation:
1. Equally inclined to the axes ⇒ slope ±1; y-intercept −1.
2. y = x − 1 ⇒ x − y − 1 = 0 and y = −x − 1 ⇒ x + y + 1 = 0.

Teacher's Note:
1. Use y = mx + c with m = ±1 and c = −1.
2. Check that both lines give y = −1 when x = 0.

Answer: (a) \( \frac{41}{7} \)

Explanation:
1. Point dividing in ratio 3 : 4 = \( \left( \frac{3(-1) + 4(2)}{7}, \frac{3(2) + 4(3)}{7} \right) = \left( \frac{5}{7}, \frac{18}{7} \right) \).
2. It lies on x + 2y = k ⇒ k = \( \frac{5}{7} + \frac{36}{7} = \frac{41}{7} \).

Teacher's Note:
1. In the section formula the ratio m : n multiplies the far point by m and the near point by n.
2. The dividing point is the only point of the segment on the line, so it must satisfy the line.

Answer: (d) None of these

Explanation:
1. \( \vec{BA} \) = (−√3, −1) and \( \vec{BC} \) = (−1, √3); their dot product is √3 − √3 = 0, so the right angle is at B.
2. The fourth vertex D = A + C − B = \( (0, 1 + \sqrt{3}) \).
3. Slope of BD = \( \frac{2 - 1 - \sqrt{3}}{\sqrt{3} + 1} = \frac{1 - \sqrt{3}}{1 + \sqrt{3}} = \sqrt{3} - 2 \).
4. BD: \( y = (\sqrt{3} - 2)x + (1 + \sqrt{3}) \), which is none of the given lines.

Teacher's Note:
1. First find which vertex has the right angle; the diagonal through B joins B to the fourth vertex D.
2. In a square (or any parallelogram) ABCD, D = A + C − B.

Answer: (c) 2 : 3

Explanation:
1. Let the ratio be k : 1. The point is \( \left( \frac{8k + 3}{k + 1}, \frac{9k - 1}{k + 1} \right) \).
2. It lies on y − x + 2 = 0: (9k − 1) − (8k + 3) + 2(k + 1) = 0 ⇒ 3k − 2 = 0 ⇒ k = 2/3.
3. So the ratio is 2 : 3.

Teacher's Note:
1. Take the ratio as k : 1 so that only one unknown appears.
2. A positive k means the line divides the segment internally.

Answer: (b) (7, 5), (−1, −1)

Explanation:
1. A point on the line can be written as (3 + h, 2 + k) with \( \frac{k}{h} = \frac{3}{4} \), i.e., k = 3h/4.
2. Distance 5: \( h^2 + \frac{9h^2}{16} = 25 \Rightarrow \frac{25h^2}{16} = 25 \Rightarrow h = \pm 4 \), k = ±3.
3. Points: (3 + 4, 2 + 3) = (7, 5) and (3 − 4, 2 − 3) = (−1, −1).

Teacher's Note:
1. A slope of 3/4 means a run of 4 for a rise of 3, which is a 3-4-5 right triangle.
2. So moving 5 units along the line changes x by ±4 and y by ±3 together.

Answer: (a) 2x − 7y − 20 = 0

Explanation:
1. x = −7 − 5y; then 3(−7 − 5y) + 2y − 5 = 0 ⇒ y = −2, x = 3.
2. Perpendicular to 7x + 2y − 5 = 0: 2x − 7y + λ = 0.
3. Through (3, −2): 6 + 14 + λ = 0 ⇒ λ = −20, so 2x − 7y − 20 = 0.

Teacher's Note:
1. Substitution is convenient when one variable has coefficient 1.
2. Check perpendicularity: slopes −7/2 and 2/7 multiply to −1.

Answer: (a) y = x, y + x = 1

Explanation:
1. The vertices are (0, 0), (1, 0), (1, 1) and (0, 1).
2. Diagonal through (0, 0) and (1, 1): y = x.
3. Diagonal through (1, 0) and (0, 1): x + y = 1.

Teacher's Note:
1. Write down the vertices first by intersecting the given lines.
2. The diagonals of a square are perpendicular: slopes 1 and −1.

Answer: (b) x − y + 12 = 0

Explanation:
1. Let D(1, 3), E(5, 7), F(−5, 7) be the mid-points of BC, CA, AB.
2. By the mid-point theorem AB ∥ DE; slope of DE = \( \frac{7 - 3}{5 - 1} = 1 \).
3. AB passes through F(−5, 7): y − 7 = x + 5 ⇒ x − y + 12 = 0.

Teacher's Note:
1. The line joining the mid-points of two sides is parallel to the third side.
2. So each side passes through one mid-point and is parallel to the line joining the other two.

Answer: (b) 2Tx − a²y + 2aT = 0

Explanation:
1. Let the line be \( \frac{x}{-a} + \frac{y}{b} = 1 \). Area = \( \frac{1}{2} \cdot a \cdot b = T \Rightarrow b = \frac{2T}{a} \).
2. So \( -\frac{x}{a} + \frac{ay}{2T} = 1 \). Multiplying by 2aT: −2Tx + a²y = 2aT.
3. Hence 2Tx − a²y + 2aT = 0.

Teacher's Note:
1. Area of the triangle cut off by a line with intercepts a and b is |ab|/2.
2. The x-intercept is already known, so only the y-intercept is to be found.

Answer: (b) 2x − y − 4 = 0

Explanation:
1. The other two vertices lie on the other diagonal, which passes through the mid-point of AC.
2. Mid-point of AC = (3, 2).
3. Line with gradient 2 through (3, 2): y − 2 = 2(x − 3) ⇒ 2x − y − 4 = 0.

Teacher's Note:
1. The diagonals of a rectangle bisect each other, so both pass through the same mid-point.
2. Unlike a square, the diagonals of a rectangle need not be perpendicular, so the gradient is given separately.

Answer: (a) 2x + y = 4

Explanation:
1. Let the line be \( \frac{x}{a} + \frac{y}{2a} = 1 \Rightarrow 2x + y = 2a \).
2. Through (1, 2): 2 + 2 = 2a ⇒ 2a = 4.
3. Hence 2x + y = 4.

Teacher's Note:
1. Write both intercepts in terms of one unknown a.
2. Check: the intercepts of 2x + y = 4 are 2 and 4.

Answer: (a) 3x − 2y = 30

Explanation:
1. Mid-point of (2, −19) and (6, 1) = (4, −9).
2. Slope of the line joining (−1, 3) and (5, −1) = \( \frac{-4}{6} = -\frac{2}{3} \), so the perpendicular slope is \( \frac{3}{2} \).
3. \( y + 9 = \frac{3}{2}(x - 4) \Rightarrow 2y + 18 = 3x - 12 \Rightarrow 3x - 2y = 30 \).

Teacher's Note:
1. "Bisects the line joining two points" means it passes through their mid-point.
2. The perpendicularity condition gives the slope from the other pair of points.

Answer: (a) \( 2x + 2y + \sqrt{2} = 0 \)

Explanation:
1. Slope of BC = \( \frac{-3 + 1}{-1 + 3} = -1 \), so the line is x + y + c = 0.
2. Distance from the origin: \( \frac{|c|}{\sqrt{2}} = \frac{1}{2} \Rightarrow c = \pm\frac{1}{\sqrt{2}} \).
3. To cut OB and OC (which lie in the third quadrant) the line must lie on the same side of O as B and C, so c > 0: \( x + y + \frac{1}{\sqrt{2}} = 0 \Rightarrow 2x + 2y + \sqrt{2} = 0 \).

Teacher's Note:
1. Two lines at the same distance from the origin and parallel to each other lie on opposite sides of it; pick the one towards B and C.
2. Points with negative x and y make x + y negative, so the line x + y = −c with c > 0 lies on their side.

Answer: (a) \( \frac{x}{x_1} + \frac{y}{y_1} = 2 \)

Explanation:
1. If the line meets the axes at (a, 0) and (0, b), the mid-point is (a/2, b/2) = (x₁, y₁).
2. So a = 2x₁, b = 2y₁ and the line is \( \frac{x}{2x_1} + \frac{y}{2y_1} = 1 \Rightarrow \frac{x}{x_1} + \frac{y}{y_1} = 2 \).

Teacher's Note:
1. The intercepts are twice the coordinates of the mid-point.
2. This result is worth remembering for quick answers.

Answer: (b) 8x − 5y + 60 = 0

Explanation:
1. Let the line meet the axes at A(a, 0) and B(0, b). The point dividing AB in the ratio 1 : 2 is \( \left( \frac{2a}{3}, \frac{b}{3} \right) \).
2. \( \frac{2a}{3} = -5 \Rightarrow a = -\frac{15}{2} \); \( \frac{b}{3} = 4 \Rightarrow b = 12 \).
3. \( \frac{x}{-15/2} + \frac{y}{12} = 1 \Rightarrow -\frac{2x}{15} + \frac{y}{12} = 1 \Rightarrow -8x + 5y = 60 \Rightarrow 8x - 5y + 60 = 0 \).

Teacher's Note:
1. The intercept is taken from the x-axis point to the y-axis point, so the ratio 1 : 2 is measured from A.
2. Check: (−5, 4) satisfies 8(−5) − 5(4) + 60 = 0.

Answer: (a) 3x − 2y = 0

Explanation:
1. Vertices: (0, 0); (1/2, 0) where 6x + y = 3 meets y = 0; (0, 1) where x + y = 1 meets x = 0; and the intersection of x + y = 1 and 6x + y = 3.
2. Subtracting, 5x = 2 ⇒ x = 2/5, y = 3/5.
3. The diagonal through the origin joins (0, 0) to \( \left(\frac{2}{5}, \frac{3}{5}\right) \): \( y = \frac{3}{2}x \Rightarrow 3x - 2y = 0 \).

Teacher's Note:
1. The vertex opposite the origin is the intersection of the two lines that do not pass through it.
2. Draw a rough sketch to see which vertices are adjacent.

Answer: (a) x − 7y + 12 = 0

Explanation:
1. The equal sides make 45° with the hypotenuse, whose slope is \( -\frac{3}{4} \).
2. \( \tan 45^\circ = \left| \frac{m + \frac{3}{4}}{1 - \frac{3m}{4}} \right| \Rightarrow \frac{4m + 3}{4 - 3m} = \pm 1 \Rightarrow m = \frac{1}{7} \) or m = −7.
3. Through (2, 2): \( y - 2 = \frac{1}{7}(x - 2) \Rightarrow x - 7y + 12 = 0 \), and y − 2 = −7(x − 2) ⇒ 7x + y − 16 = 0.
4. Of these, x − 7y + 12 = 0 is among the options.

Teacher's Note:
1. In an isosceles right triangle the base angles are 45°, so each leg makes 45° with the hypotenuse.
2. The two legs found are perpendicular: (1/7)(−7) = −1, which confirms the working.

Answer: (a) 52x + 89y + 519 = 0

Explanation:
1. Since AB = AC, AB and AC make equal angles with BC (slope \( \frac{3}{4} \)), on opposite sides.
2. For AB (slope −4): \( \frac{-4 - \frac{3}{4}}{1 + (-4)\frac{3}{4}} = \frac{19}{8} \). For AC (slope m): \( \frac{m - \frac{3}{4}}{1 + \frac{3m}{4}} = -\frac{19}{8} \).
3. ⇒ 8(4m − 3) = −19(4 + 3m) ⇒ 89m = −52 ⇒ \( m = -\frac{52}{89} \).
4. AC: \( y + 7 = -\frac{52}{89}(x - 2) \Rightarrow 52x + 89y + 519 = 0 \).

Teacher's Note:
1. In an isosceles triangle the base angles are equal, so the equal sides are equally inclined to the base.
2. Use the tan formula with opposite signs for the two sides; the same sign would give back AB itself.